Compute $\sum_{k\geq0}\frac{2^k}{5^{2^k}+1}$

Question

Problem Statement:

Compute: $$\sum_{k\geq0}\frac{2^k}{5^{2^k}+1}$$

First we note that $\sum_{k\geq0}\frac{2^k}{5^{2^k}+1}< \sum_{k\geq0}\frac{2^k}{5^{2^k}}< \sum_{k\geq0}\frac{2^k}{5^{k}}< \infty$, so that this sum converges. We can now do the following manipulations: \begin{align} \sum_{k\geq0}\frac{2^k}{5^{2^k}+1}&= \sum_{k\geq0}\frac{\frac{2^k}{5^{2^k}}}{1+ \frac{1}{5^{2^k}}}\\ &= \sum_{k\geq0}\frac{2^k}{5^{2^k}}\left(\sum_{j\geq0}\frac{(-1)^j}{(5^{2^k})^j}\right)\quad \quad \text{geometric series with $r = -\frac{1}{5^{2^k}}$}\\ &= \sum_{k\geq0}2^k\left(\sum_{j\geq0}\frac{(-1)^j}{(5^{2^k})^{j+1}}\right)\\ &=\sum_{j\geq0}(-1)^j\left(\sum_{k\geq0}\frac{2^k}{(5^{j+1})^{2^k}}\right) \quad \quad \text{swap sums due to convergence} \end{align}

I was hoping that after swapping sums the new summation would be easier to evaluate. However, this doesn't appear to be so. Any hints would be greatly appreciated.

Answer 1

This is a beautiful sum. Note:

$$\frac 1 {n-1} -\frac1{n+1} =\frac {2}{n^2-1}$$

Thus: $$\begin{align}\frac 1 {n-1} -\frac1{n+1}-\frac {2}{n^2+1} &=\frac {2}{n^2-1}-\frac {2}{n^2+1}\\& = \frac {2^2}{n^{2^2}-1}\\\frac 1 {n-1} -\frac1{n+1}-\frac {2}{n^2+1}-\frac {2^2}{n^{2^2}+1} &= \frac {2^2}{n^{2^2}-1}-\frac {2^2}{n^{2^2}+1}\quad\quad\quad\ \ \\&= \frac {2^3}{n^{2^3}-1}\end{align}$$

Now complete the pattern.

EDIT: After a brief discussion with @Integrand, we concluded that this is much easier to explain with a telescoping sum, and by observing:

$$\frac {2^k} {n^{2^k}+1} = \frac{2^k} {n^{2^k}-1} - \frac{2^{k+1}} {n^{2^{k+1}}-1}$$

Now

$$\begin{align}\sum_{k=0}^N\frac {2^k} {n^{2^k}+1} &= \sum_{k=0}^N\left(\frac{2^k} {n^{2^k}-1} - \frac{2^{k+1}} {n^{2^{k+1}}-1}\right)\\&= \sum_{k=0}^N\frac{2^k} {n^{2^k}-1} - \sum_{k=1}^{N+1}\frac{2^{k}} {n^{2^{k}}-1}\\&=\frac{1}{n-1}-\frac{2^{N+1}}{n^{2^{N+1}}-1}\end{align}$$

and we see that $\dfrac{2^{N+1}}{n^{2^{N+1}}-1} \to 0$ as $N\to \infty$.